# Boom Town

Alignments to Content Standards: F-LE.A.2

The population of Pittsburgh was about 12,600 in 1820. Between 1820 and 1840 the population grew exponentially, increasing by about 70% each decade.

1. Construct an exponential function in the form $f(t) = a b^t$ that models the population $t$ decades after 1820.
2. According to your model, what was the population of Pittsburgh in 1825? What about 1839?
3. According to your model, by what percentage did the population increase between 1826 and 1836?
4. Explain the answer to part (c) in terms of the growth factor $b$.

## IM Commentary

The purpose of this task is to give students experience working with simple exponential models in situations where they must evaluate and interpret them at non-integer inputs. Although not a proof, this task reinforces the understanding behind standard F-LE.A.1a, that exponential functions grow by equal factors over equal intervals, even when those intervals do not have integer endpoints.

Through the course of this task, one ends up evaluating the expression $12,600(1.7)^t$ for several values of $t$. Many calculator platforms offer the ability to enter a function one time and then quickly evaluate that function at different values. This could be a nice opportunity for students to use appropriate tools strategically MP.5.

## Solution

1. The population $0$ decades after 1820 is 12,600, so the initial value is $f(0) = ab^0 = a \cdot 1 = a = 12,600$. The population grows by 70% each decade, and so gets multiplied by $1 + 0.7 = 1.7$ each decade. So $b = 1.7$ and $f(t) = 12,600(1.7)^t$.
2. The year 1825 is halfway through the first decade, so corresponds to $t = 0.5$. The population predicted by the model in 1825 is $f(0.5) = 12,600(1.7)^{0.5} \approx 16428$, or about $16,400$. The year 1839 is 19 years after 1720, so corresponds to $t = 1.9$ decades. The predicted population in that year is $f(1.9) = 12,600(1.7)^{1.9} \approx 34532$, or about 34,500.
3. Between 1826 ($t=0.6$) and 1836 ($t=1.6$) the percent change in population is $\frac{f(1.6)-f(0.6)}{f(0.6)} = 0.7$, so the percentage increase is 70%.
4. Exponential functions grow by constant factors over constant intervals, no matter which interval we look at. We know that the population grows by 70% each decade, so we would also expect 70% growth during the decade from 1826 to 1836.